LeetCode #3335 — MEDIUM

Total Characters in String After Transformations I

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

You are given a string s and an integer t, representing the number of transformations to perform. In one transformation, every character in s is replaced according to the following rules:

If the character is 'z', replace it with the string "ab".
Otherwise, replace it with the next character in the alphabet. For example, 'a' is replaced with 'b', 'b' is replaced with 'c', and so on.

Return the length of the resulting string after exactly t transformations.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: s = "abcyy", t = 2

Output: 7

Explanation:

First Transformation (t = 1):
- 'a' becomes 'b'
- 'b' becomes 'c'
- 'c' becomes 'd'
- 'y' becomes 'z'
- 'y' becomes 'z'
- String after the first transformation: "bcdzz"
Second Transformation (t = 2):
- 'b' becomes 'c'
- 'c' becomes 'd'
- 'd' becomes 'e'
- 'z' becomes "ab"
- 'z' becomes "ab"
- String after the second transformation: "cdeabab"
Final Length of the string: The string is "cdeabab", which has 7 characters.

Example 2:

Input: s = "azbk", t = 1

Output: 5

Explanation:

First Transformation (t = 1):
- 'a' becomes 'b'
- 'z' becomes "ab"
- 'b' becomes 'c'
- 'k' becomes 'l'
- String after the first transformation: "babcl"
Final Length of the string: The string is "babcl", which has 5 characters.

Constraints:

1 <= s.length <= 10⁵
s consists only of lowercase English letters.
1 <= t <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given a string s and an integer t, representing the number of transformations to perform. In one transformation, every character in s is replaced according to the following rules: If the character is 'z', replace it with the string "ab". Otherwise, replace it with the next character in the alphabet. For example, 'a' is replaced with 'b', 'b' is replaced with 'c', and so on. Return the length of the resulting string after exactly t transformations. Since the answer may be very large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Math · Dynamic Programming

Example 1

"abcyy"
2

Example 2

"azbk"
1

Step 02

Core Insight

What unlocks the optimal approach

Maintain the frequency of each character.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3335: Total Characters in String After Transformations I
class Solution {
    public int lengthAfterTransformations(String s, int t) {
        final int mod = (int) 1e9 + 7;
        int[][] f = new int[t + 1][26];
        for (char c : s.toCharArray()) {
            f[0][c - 'a']++;
        }
        for (int i = 1; i <= t; ++i) {
            f[i][0] = f[i - 1][25] % mod;
            f[i][1] = (f[i - 1][0] + f[i - 1][25]) % mod;
            for (int j = 2; j < 26; j++) {
                f[i][j] = f[i - 1][j - 1] % mod;
            }
        }

        int ans = 0;
        for (int j = 0; j < 26; ++j) {
            ans = (ans + f[t][j]) % mod;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3335: Total Characters in String After Transformations I
func lengthAfterTransformations(s string, t int) int {
	const mod = 1_000_000_007
	f := make([][]int, t+1)
	for i := range f {
		f[i] = make([]int, 26)
	}

	for _, c := range s {
		f[0][c-'a']++
	}

	for i := 1; i <= t; i++ {
		f[i][0] = f[i-1][25] % mod
		f[i][1] = (f[i-1][0] + f[i-1][25]) % mod
		for j := 2; j < 26; j++ {
			f[i][j] = f[i-1][j-1] % mod
		}
	}

	ans := 0
	for j := 0; j < 26; j++ {
		ans = (ans + f[t][j]) % mod
	}
	return ans
}

# Accepted solution for LeetCode #3335: Total Characters in String After Transformations I
class Solution:
    def lengthAfterTransformations(self, s: str, t: int) -> int:
        f = [[0] * 26 for _ in range(t + 1)]
        for c in s:
            f[0][ord(c) - ord("a")] += 1
        for i in range(1, t + 1):
            f[i][0] = f[i - 1][25]
            f[i][1] = f[i - 1][0] + f[i - 1][25]
            for j in range(2, 26):
                f[i][j] = f[i - 1][j - 1]
        mod = 10**9 + 7
        return sum(f[t]) % mod

// Accepted solution for LeetCode #3335: Total Characters in String After Transformations I
/**
 * [3335] Total Characters in String After Transformations I
 */
pub struct Solution {}

// submission codes start here

const MOD: i32 = 1_000_000_007;

impl Solution {
    pub fn length_after_transformations(s: String, t: i32) -> i32 {
        let mut map = vec![0; 26];

        for c in s.bytes() {
            map[(c - b'a') as usize] += 1;
        }

        let mut result = s.bytes().len() as i32;

        for _ in 0..t {
            let z_count = map[25];

            for c in (0..25).rev() {
                if map[c] != 0 {
                    map[c + 1] = (map[c + 1] + map[c]) % MOD;
                    map[c] = 0;
                }
            }

            if z_count != 0 {
                map[0] = (map[0] + z_count) % MOD;
                map[1] = (map[1] + z_count) % MOD;
                result = (result + z_count) % MOD;

                map[25] = (map[25] + MOD - z_count) % MOD;
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3335() {
        assert_eq!(
            7,
            Solution::length_after_transformations("abcyy".to_string(), 2)
        );
        assert_eq!(
            5,
            Solution::length_after_transformations("azbk".to_string(), 1)
        );
        assert_eq!(
            79033769,
            Solution::length_after_transformations("jqktcurgdvlibczdsvnsg".to_string(), 7517)
        );
    }
}

// Accepted solution for LeetCode #3335: Total Characters in String After Transformations I
function lengthAfterTransformations(s: string, t: number): number {
    const mod = 1_000_000_007;
    const f: number[][] = Array.from({ length: t + 1 }, () => Array(26).fill(0));

    for (const c of s) {
        f[0][c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
    }

    for (let i = 1; i <= t; i++) {
        f[i][0] = f[i - 1][25] % mod;
        f[i][1] = (f[i - 1][0] + f[i - 1][25]) % mod;
        for (let j = 2; j < 26; j++) {
            f[i][j] = f[i - 1][j - 1] % mod;
        }
    }

    let ans = 0;
    for (let j = 0; j < 26; j++) {
        ans = (ans + f[t][j]) % mod;
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(t × |\Sigma|)

Space

O(t × |\Sigma|)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.